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Firstly, I want to show $(\Bbb R , +)$ is isomorphic to $\Bbb R^2$ (under component wise addition).

In this I used the mapping $\phi_1$ : $\Bbb R^2 \rightarrow \Bbb R$

Where, $\phi_1(a,b) = 2^a3^b$

And then , show $\Bbb R^2$ is isomorphic to $(\Bbb C , +)$.

In this I used the mapping

$\phi_2$ : $\Bbb R^2 \rightarrow \Bbb C$

Where, $\phi_2(a,b) = a+ib$

And then, use the transitive property of isomorphic groups to show $(\Bbb R , +)$ is isomorphic to $(\Bbb C , +)$

Now ,my doubts are that, whether the mappings i used are correct or not. Does this mappings makes the two groups isomorphic?

Please, correct me wherever I am wrong?

Edit: I see that $\phi_1$ is not bijective. So , their is no way to prove ,$\phi_1$ : $\Bbb R^2 \rightarrow \Bbb R$ is an isomorphism by just defining a mapping or without using the concept of metric space!

Rkb
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    $\phi_1$ is not even a morphism, it's not injective and it's not surjective : very far from an isomorphism – Maxime Ramzi May 15 '19 at 10:49
  • @Max Thanks for your help. So can you define a mapping for this if possible & what about $\phi_2$ ? – Rkb May 15 '19 at 10:52
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    $\phi_2$ is alright (if you're not sure, you should prove it !). However, defining $\phi_1$ is going to be complicated : we can prove that it exists but it's hard (or impossible) to actually make it explicit. The proof I know requires the axiom of choice, whereas the result does not hold without assuming any form of it, so there will be some “non-explicit part“ – Maxime Ramzi May 15 '19 at 10:59
  • Your map $\phi_1$ is an injective map $\Bbb N^2\to\Bbb N,$ but isn't even surjective there. It is still injective (but not surjective) if you extend to $\Bbb Z^2\to\Bbb Q,$ or $\Bbb Q^2\to\Bbb R.$ Unfortunately, you lose injectivity when you extend it to $\Bbb R^2\to\Bbb R.$ – Cameron Buie May 15 '19 at 11:07
  • I see that , most of answers to this question requires metric space. – Rkb May 15 '19 at 11:08
  • @Arthur And also i am not familiar with metric space, as i am just a begginer. So, their is no method of proving this isomorphism in $\phi_1$ by just simply defining a function. – Rkb May 15 '19 at 11:12
  • @CameronBuie yeah i noticed it as the above person said. – Rkb May 15 '19 at 11:13
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    I do not know any answer to this that uses metric space. I do know one using vector space. – GEdgar May 15 '19 at 12:24
  • @GEdgar Thanks for your help, but unfortunately the question has been closed , so i can't get any more opinions – Rkb May 15 '19 at 13:39
  • Did you mean "vector space" rather than "metric space"? Otherwise your question is not a duplicate and should be reopened. – user1729 May 15 '19 at 16:22
  • @user1729 sorry but i don't know either of them. What i want ,is just to define a bijective mapping from $\Bbb R^2$ to $\Bbb R$ – Rkb May 15 '19 at 16:45
  • I think metric spaces are totally irrelevant here, you can as well ask about a proof which uses, say, the probability theory. – Moishe Kohan May 16 '19 at 00:08

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